Given file.txt that looks like this:
line_ some text
line_ some text
line_ some text
How can I number the lines like this with Bash:
line_1 some text
line_2 some text
line_3 some text
Here is my idea so far, but it doesn't work:
#!/bin/bash
var = 1
cat ./file.txt
while read line; do
sed "s/line_/line_(( var++ ))/"
done < ./file.txt
You can use awk:
awk '$1=$1 FNR' <file>
$1 is the first word of each record (line in this case). FNR is the input record number (line number in this case).
This command is replacing the first word of each line with the first word + line-number of each line.
Here is how to do it in perl:
perl -pe 'print "line_$. $_";' [file(s)]
If you want to send it to a file:
perl -pe 'print "line_$. $_";' [input_file] > [output_file]
line_ some text -> line_1 line_ some text. Is it correct?
line_. Since you just want to add the line number after the first occurrence of an underscore, you can just use perl -ne 's/_/$&$./' input_file > output_file
-pe. -ne prints nothing.
It looks to me like you're well covered here, but what the hell:
sed -et -e's/_ /\n/;P;=;D' <in | paste -'d_ ' - - - >out
sed's not exactly a whiz in the math department, but sometimes friends help it cheat, and so it gets by.
That transforms...
line_ some text
line_ some text
line_ some text
...to...
line_1 some text
line_2 some text
line_3 some text
Well, you are quite new to Shell/Bash it seems, so for educational reasons, here is a solution for you, that covers your technology of choice (http://mywiki.wooledge.org/BashFAQ/001, http://mywiki.wooledge.org/BashGuide/Parameters#Parameter_Expansion)
var=1
while read -r Line; do
printf "%s\n" "${Line/line_/line_$var}" >> ./file-new.txt
let var++
done < ./file.txt
